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Clash Royale CLAN TAG #URR8PPP up vote 4 down vote favorite 1 My lecture notes state the following: When we were dealing with first order equations we saw that a differential operator of the form, $$pfracpartialpartialx + qfracpartialpartialy$$ Led to the characteristic equations $$fracdxdt = p, fracdydt = q$$ The solution of these ODE in turn implied that the solution would be constant along characteristics of the form $qx − py = constant$. Can someone please demonstrate this? differential-equations pde differential-operators characteristics share | cite | improve this question asked Aug 11 at 4:51 handler's handle 150 8 2 Are $p,q$ constants? Usually in the treatment of first order PDE one has $p=frac∂u∂x$, $q=frac∂u∂y$. – Lutz...

Clash Royale CLAN TAG #URR8PPP up vote 0 down vote favorite $f(x) = int_1^infty frac2sqrt2pie^-frac12x^2 dx$ find $P(X > 1)$ This is $X$ ~ $Norm(0, 1)$. $P(X > 1) = 1 - P(X leq 1) = 1 - 2 phi(1) = 1-2(1-phi(-1)) = 1 - 2(1-0.1587) = -0.6826$. Yikes. Negative number. What am I doing wrong? probability share | cite | improve this question edited Aug 11 at 5:51 Did 242k 23 208 443 asked Aug 11 at 4:02 Bas bas 396 11 First your $f(x)$ has an error in that there should be no $2$ in the numerator of $frac2sqrt2pi$, further, expression for $f$ is independent of $x$ and unnecessary. It's not clear why you are manipulating the expression for $P(X>1)$ here since you're going to need to resort to numerically integrating the Gaussian at some point anyway. In my opinion, the easiest way to do this problem is to depend on the Empirical rule to conclude that $P(0leq X leq 1) approx 0.6827/2$ and so $P(X>1) approx 0.5- 0....